Calling java script in PHP

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  • Dormilich
    replied
    Originally posted by dheerajjoshim
    Its a string...
    it ain’t even a string, it’s NULL because you assign a set of invalid characters. (strings are wrapped in ' (single quotation mark or apostrophe) or " (double quotation mark) not backticks (these are used in SQL).

    but of course you can use PHP to print JS code…

      Leave a comment:

    • Dheeraj Joshi
      replied
      Its a string...

      I am sorry..

      So can i use java script in PHP?

      Regards
      Dheeraj Joshi

        Leave a comment:

      • Dormilich
        replied
        Originally posted by dheerajjoshim
        Code:
        $res = `php myfile.php`;
        how is that an integer?

          Leave a comment:

        • Dheeraj Joshi
          replied
          $res is integer by the way.

          So isnt there a technique to pop up a window in PHP code?

          Regards
          Dheeraj Joshi

            Leave a comment:

          • Dormilich
            replied
            Originally posted by dheerajjoshim
            Technically it should pop up a window of alert.
            No, it should give "No choice". if you do var_dump($res); you’ll see why. besides, even if $res would be a string it still would match the default chioce.

            Originally posted by dheerajjoshim
            this will give value 1.
            how do you come to that conclusion?

              Leave a comment:

            • Dheeraj Joshi
              started a topic Calling java script in PHP

              Calling java script in PHP

              I have this code.


              Code:
              <html>
<body>
<?php

$res = `php myfile.php`;

switch($res)
{
	case 1: "<scriptlanguage=javascript> 
		alert('Blah Blah')
		</script>";
		break;
	case 2: echo "Choice 2";
		break;
	case 3: echo "Choice 3";
		break;
	default:echo "No choice";
		break;
}

?>
</body>
</html>
              Technically it should pop up a window of alert.

              But it wont.
              Code:
              echo $res;
              this will give value 1.

              Whats wrong with it.?

              Regards
              Dheeraj Joshi
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